OEInscribed means that the edges of the shapes are touching. The
radius of the circle is 3, which means that PO is 3. If Z were at the other
end of the diameter from P, this problem would be easy and the answer
would be 6, right? But Z is beyond the edge of the circle, which means
that PZ is a little more than 6. Let’s stop there for a minute and glance
at the answer choices. We can eliminate anything that’s “out of the
ballpark”—in other words, any answer choice that’s less than 6, equal
to 6 itself, or a lot more than 6. Remember when we told you to
memorize a few of those square roots?
Let’s use them:
(A) Exactly 6? Nope.
(B) That’s 1.4 × 3, which is 4.2. Too small.
(C) That’s 6 + 1.4, or 7.4. Not bad. Let’s leave that one in.
(D) That’s 3 + 1.7, or 4.7. Too small.
(E) That’s (3 × 1.4) + 3, which is 4.2 + 3, or 7.2. Not bad. Let’s
So we eliminated three choices with Ballparking. We’re left with (C)
and (E). You could take a guess here if you had to, but let’s do a little
more geometry to find the correct answer.
Because this circle is inscribed in the square, the diameter of the circle
is the same as a side of the square. We already know that the diameter
of the circle is 6, so that means that ZY, and indeed all the sides of the
square, are also 6. Now, if ZY is 6, and XY is 6, what’s XZ, the diagonal
of the square? Well, XZ is also the hypotenuse of the isosceles right
triangle XYZ. The hypotenuse of a right triangle with two sides of 6 is \(6 \sqrt{2}\) . That’s approximately 6 × 1.4, or 8.4.
The question is asking for PZ, which is a little less than XZ. It’s
somewhere between 6 and 8.4. The pieces that aren’t part of the
diameter of the circle are equal to 8.4 − 6, or 2.4. Divide that in half to
get 1.2, which is the distance from the edge of the circle to Z. That
means that PZ is 6 + 1.2, or 7.2. Check your remaining answers: Choice
(C) is 7.4, and (E) is 7.2. Bingo! The answer is (E).
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